Gravitational Attraction Question

In summary: E = (1.74x10^6m)2(1.67 N)/(6.67x10^-11 Nm2/kg2)(7.35x10^22 kg) mM = (5.57x10^23 kg)2(1.62 N)/(6.38x10^6 m2/kg2)(3.84x10^8 m)2...this will give you the equation for the net force on the moon:Fnet = Fg of Earth - Fg of the MoonFnet = (9.80 N - 1.62 N)Now that you have the equation for the net
  • #36
wilson_chem90 said:
so would it just be X+X = D {Me/Mm}
No. Realize that:

X = (D-X) {Me/Mm} →

X = D{Me/Mm} - X{Me/Mm}
 
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  • #37
oh okay, yeah i remember that now, is that the final equation though?
 
  • #38
What do you mean by "final"? You still have to solve for X.
 
  • #39
well i have to isolate the other x right, but the only think i can think of is it would be X^2 = D{Me/Mm} - {Me/Mm}
 
  • #40
wilson_chem90 said:
well i have to isolate the other x right, but the only think i can think of is it would be X^2 = D{Me/Mm} - {Me/Mm}
No. You must isolate X, but you don't end up with X^2.

How would you solve this one?

x = a -bx (a and b are just constants)

It's the same basic equation.
 
  • #41
would it be x = a/b?
 
  • #42
wilson_chem90 said:
would it be x = a/b?
No. Do it step by step:

x = a - bx

group the x terms on the left:
x + bx = a

factor out the x:
x (1 + b) = a

isolate the x by dividing both sides by (1 + b):
x = a/(1 + b)

done!
 
  • #43
YES! finally, thank you so much. i appreciate your time and patience with me
 
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